Found it: Let P be the midpoint of GE; Q the midpoint of EJPEQB is a parallelogram (PB is midpoint segment in Triangle EGJ)Triangle FPB congruent with Triangle BQHHence, FB=BH<FBP+<FBH+<HBQ+<BQE=180 (cointerior angles in p'gram) I<HBQ+<BQE+<QHB=90 (adds up to 180 in triangle) II